Difficulty: Intermediate | Prerequisites: Lewis structures, electronegativity, acid-base basics (Ch. 1-4)
This chapter is where organic chemistry stops being about naming molecules and starts being about what they do. Addition reactions are the signature moves of alkenes and alkynes, the unsaturated hydrocarbons whose pi bonds make them reactive. If you skipped or forgot acid-base behaviour from earlier chapters, go back now, because every mechanism here hinges on electron-pair movement. Mastering these reactions gives you the toolkit for predicting products, understanding regioselectivity, and recognising when rearrangements can (or cannot) occur.
Alkenes react by addition: a pi bond breaks, two new sigma bonds form, and the result is exothermic. Different reagents (HX, H₂O/H⁺, Br₂, Hg(OAc)₂, BH₃) follow different mechanistic paths, but they all start with the pi bond acting as a nucleophile. Regioselectivity (which carbon gets which atom) depends on the mechanism.
Addition reaction
A reaction in which atoms or groups add across a double (or triple) bond, breaking one pi bond and one sigma bond in the reagent and forming two new sigma bonds. The process is exothermic because bonds formed are stronger than bonds broken.
In simple terms, you are "filling in" the double bond with new atoms.
Electrophile
An electron-poor species (or region) that accepts an electron pair. In addition reactions, the electrophile is the species attracted to the pi electrons of the alkene. Common examples: H⁺, Br₂, Hg(OAc)₂.
Think of it as the "electron-lover" that goes after the electron-rich pi bond.
Nucleophile
An electron-rich species that donates an electron pair to an electrophile. In addition reactions, the pi bond itself acts as the nucleophile in the first step. Common examples: H₂O, Br⁻, OH⁻, BH₃.
Think of it as the "nucleus-lover" that brings electrons to the party.
Markovnikov's rule
In the addition of HX to an alkene, the hydrogen adds to the carbon with more hydrogens (fewer substituents), and the X group adds to the carbon with more substituents. This follows from formation of the more stable carbocation intermediate.
In simple terms, "the rich get richer": the carbon that already has more alkyl groups picks up the new substituent.
Carbocation
A positively charged carbon intermediate. Stability order: tertiary > secondary > primary > methyl. Carbocations are subject to rearrangement (hydride and methyl shifts) to reach greater stability.
Bromonium ion
A three-membered cyclic intermediate formed when Br₂ adds to an alkene. The bromine bridges both carbons, carrying a formal positive charge. This intermediate avoids a full carbocation forming, so no rearrangement occurs.
Think of it as bromine sitting on top of both carbons like a bridge, blocking one face of the molecule.
Halohydrin
The product of adding a halogen and a hydroxyl group across a double bond (e.g. bromohydrin from Br₂/H₂O). Water, being far more abundant in solution, attacks the bromonium ion instead of bromide.
Vicinal dihalide
A compound with two halogen atoms on adjacent (neighbouring) carbons. Formed by halogen addition (e.g. Br₂) to an alkene.
Geminal dihalide
A compound with two halogen atoms on the same carbon.
All electrophilic additions to alkenes share two fundamental steps:
Electron-pair donation from the pi bond to the electrophile. The pi bond acts as a nucleophile, attacking the electrophilic species. The electrophile adds to the carbon with more hydrogens (fewer substituents).
Nucleophilic atom donates an electron pair to carbon. A nucleophile attacks the carbocation (or bridged intermediate) at the more substituted carbon.
The overall energy balance: bonds broken < bonds formed, so addition reactions are exothermic (ΔH < 0).
Water alone is not a strong enough acid to protonate an alkene. Sulfuric acid (H₂SO₄) acts as a catalyst, providing H⁺.
The proton adds to the less substituted carbon (Markovnikov), generating the more stable carbocation.
Water then attacks the carbocation as a nucleophile.
A final deprotonation gives the alcohol product.
Because a carbocation intermediate forms, rearrangements (hydride shifts, methyl shifts) are possible.
Br₂ approaches the alkene. The pi electrons attack one bromine, pushing the other away as Br⁻.
Instead of forming an open carbocation, bromine bridges both carbons, creating a cyclic bromonium ion.
The bromonium ion carries a partial positive charge distributed across both carbons, but charge density is greater on the more substituted carbon.
Br⁻ attacks from the opposite face (anti addition), giving a vicinal dihalide.
Key point: no carbocation forms, so no rearrangement occurs.
HBr adds across the double bond following Markovnikov's rule.
A carbocation intermediate forms, so hydride and methyl shifts can occur to give a more stable cation before the bromide attacks.
In the presence of water, the bromonium ion intermediate is attacked by water rather than Br⁻, because water is present in much higher concentration (it is the solvent).
Water attacks the more substituted carbon of the bromonium ion.
The product is a bromohydrin: Br on one carbon, OH on the adjacent carbon (anti addition).
General addition:
Alkene + A-B → product with two new sigma bonds
Broken: 1 pi bond + 1 sigma bond (in A-B)
Formed: 2 sigma bonds
Condition for exothermic: energy of bonds formed > energy of bonds broken
Students often think the bromonium ion is a true carbocation. It is not. The bridged bromine prevents the kind of open cation that allows rearrangement.
Students frequently confuse vicinal and geminal dihalides. Vicinal means the halogens are on adjacent carbons; geminal means they are on the same carbon.
A common error is assuming water is acidic enough to protonate an alkene on its own. It is not. A strong acid catalyst (H₂SO₄) is required for acid-catalysed hydration.
Students sometimes forget that halohydrin formation gives anti addition, just like bromination, because both go through the same bromonium ion intermediate.
⚠️ Know the two basic steps of every electrophilic addition. This is the framework that ties every reaction in this chapter together.
⚠️ Be able to predict when rearrangements can and cannot occur. Carbocation intermediate = rearrangement possible. Bromonium ion or concerted mechanism = no rearrangement.
⚠️ Distinguish between Markovnikov and anti-Markovnikov products. This is heavily tested.
⚠️ Understand why water attacks more than bromide in halohydrin formation (concentration, not nucleophilicity).
True or False: Addition reactions are endothermic.
Fill in the blank: In the bromonium ion intermediate, no ______ occurs because there is no open carbocation.
True or False: In halohydrin formation, bromide ion is the nucleophile that attacks the bromonium ion.
Fill in the blank: A vicinal dihalide has halogens on ______ carbons.
True or False: H₂SO₄ acts as a catalyst, not a reactant, in acid-catalysed hydration.
Answers: 1. False (exothermic). 2. Rearrangement. 3. False (water attacks, because it is present in far greater concentration). 4. Adjacent. 5. True.
Q: What are the two basic steps shared by all electrophilic addition reactions?
A: (1) The pi bond donates an electron pair to the electrophile. (2) A nucleophile donates an electron pair to the resulting carbocation or electrophilic carbon.
Q: Why does a bromonium ion intermediate prevent carbocation rearrangement?
A: The bromine atom bridges both carbons in a three-membered ring, so no carbon bears a full positive charge. Without an open carbocation, there is no driving force for a hydride or methyl shift.
Q: Predict the major product of HBr addition to 2-methylpropene (isobutylene).
A: 2-bromo-2-methylpropane. The proton adds to the terminal (less substituted) carbon, forming a tertiary carbocation, and bromide attacks that tertiary carbon.
Q: In the formation of a bromohydrin from cyclohexene with Br₂/H₂O, which nucleophile attacks the bromonium ion, and why?
A: Water attacks, not bromide. Water is the solvent and is present in much higher concentration than bromide, so it is far more likely to encounter the bromonium ion.
Q: A student claims that acid-catalysed hydration of 3-methyl-1-butene gives only 3-methyl-2-butanol. Is this correct? Explain.
A: It is plausible. Protonation of C1 gives a secondary carbocation at C2, which can rearrange via hydride shift to a more stable tertiary carbocation at C3. Water attacks C3, giving 2-methyl-2-butanol as the major product, not 3-methyl-2-butanol. The student has not accounted for carbocation rearrangement.
This material connects directly to Chapter 4 (stereochemistry), because anti addition in bromination and halohydrin formation produces specific stereochemical outcomes. Understanding carbocation stability from earlier chapters is essential for predicting Markovnikov selectivity and rearrangements here.
These electrophilic addition patterns reappear in later chapters on reactions of alkynes, dienes, and aromatic compounds. The concept of the nucleophile/electrophile pair is the single most reused idea in organic chemistry.
Electrophilic addition, Markovnikov addition, anti-Markovnikov, carbocation rearrangement, hydride shift, methyl shift, bromonium ion, cyclic halonium ion, halohydrin, bromohydrin, vicinal dihalide, geminal dihalide, acid-catalysed hydration, alkene reactions, regioselectivity, anti addition, syn addition, pi bond nucleophile, organic chemistry chapter 5